> [!META]- Inline Metadata
> [tags:: #career #concepts/programming/linked-list]
> [project group:: Neetcode 150]
> [difficulty:: easy]
> [status:: to begin]
> [link:: https://leetcode.com/problems/reverse-linked-list/]
> [up:: [[Leetcode MOC]]]
## Problem
Given the `head` of a singly linked list, reverse the list, and return _the reversed list_.
**Example 1:**
**Input:** head = [1,2,3,4,5]
**Output:** [5,4,3,2,1]
**Example 2:**
**Input:** head = [1,2]
**Output:** [2,1]
**Example 3:**
**Input:** head = []
**Output:** []
**Constraints:**
- The number of nodes in the list is the range `[0, 5000]`.
- `-5000 <= Node.val <= 5000`
## Solution
### Mine
#### Python
#### Rust
### LC's
## Things I Learned
## Scratchpad
We should be able to do this in one pass.
Maybe something like while head.next:
next.next = head
head = head.next?
### Neetcode Video
#### Iterative Solution
[[Two Pointer or Iterator Algorithm]] - prev (set to null), curr points at first.
![[Pasted image 20230110204003.png]]
```python
def ... :
prev, curr = None, head
while curr:
nxt = curr.next # Temp holder for next
curr.next = prev # Set next attribute of current pointer to prev Node (reverse)
prev = curr # Update prev pointer to point to current pointer
curr = nxt # Update curr to point to what was originally curr.next
```
T $O(n)$, M $O(1)$
#### Recursive solution
Takes M $O(N)$ because of recursive call
```python
def ...:
if not head:
return None # base case
new_head = head
if head.next:
new_head = self.reverseList(head.next)
head.next.next = head # Reverse link between next node and head
head.next = None
return new_head
```